# Question 3acdc

Feb 27, 2017

Here's how you can do that.

#### Explanation:

The idea here is that you need to work with a sample of this solution and figure out how many moles of propanol it contains.

As you know, molality is defined as the number of moles of solute present for every $\text{1 kg}$ of solvent. In your case, propanol is the solute and water is the solvent.

To make the calculations easier, let's pick a sample of solution that contains exactly

1 color(red)(cancel(color(black)("kg"))) * (10^3"g")/(1color(red)(cancel(color(black)("kg")))) = 10^3"g"

of water. You know that water has a molar mass of ${\text{18.015 g mol}}^{- 1}$, which means that this solution will contain

10^3 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "55.51 moles H"_2"O"#

Now, the mole fraction of propanol in this solution is defined as the ratio between the number of moles of propanol, let's say ${n}_{p}$, and the total number of moles present in solution.

In your case, this mole fraction is equal to

$\left({n}_{p} \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles"))))/((n_p + 55.51)color(red)(cancel(color(black)("moles}}}}\right) = 0.1538$

This means that you have

${n}_{p} = \left({n}_{p} + 55.51\right) \cdot 0.1538$

$\left(1 - 0.1538\right) \cdot {n}_{p} = 55.51 \cdot 0.1538$

$0.8462 \cdot {n}_{p} = 8.5374 \implies {n}_{p} = \frac{8.5374}{0.8462} = 10.09$

Therefore, you know that a solution of propanol that contains $\text{1 kg}$ of water and in which the solute has a $0.1538$ mole fraction contains $10.09$ moles of propanol.

You can thus say that the molality of the solution is equal to

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{\text{molality = 10.09 mol kg}}^{- 1}}}}$

The answer is rounded to four sig figs, the number of sig figs you have for the mole fraction of propanol.